For the equilibrium pcl5(g) pcl3(g) + cl2(g), kc = 2.0 × 101 at 240°c. if pure pcl5 is placed in a 1.00-l container and allowed to come to equilibrium, and the equilibrium concentration of pcl3(g) is 0.27 m, what is the equilibrium concentration of pcl5(g)?

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

Explanation:

for the reaction

PCl₅(g) → PCl₃(g) + Cl₂(g)

where

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M

and [A] denote concentrations of A

if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M

therefore

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M

[PCl₅]  =  0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M

[PCl₅]  = 3.64*10⁻³ M

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M